Find the standard form of the equation of the ellipse with the given characteristics.
Foci: $$( - 5 , - 1 ) , ( - 5,5 )$$ ,endpoints of the major axis: $$( - 5 , - 5 ) , ( - 5,9 )$$
A) $\frac { ( x - 5 ) ^ { 2 } } { 40 } + \frac { ( y + 2 ) ^ { 2 } } { 49 } = 1$
B) $\frac { ( x + 5 ) ^ { 2 } } { 40 } + \frac { ( y - 2 ) ^ { 2 } } { 49 } = 1$
C) $\frac { ( x + 5 ) ^ { 2 } } { 49 } + \frac { ( y - 2 ) ^ { 2 } } { 40 } = 1$
D) $\frac { ( x - 2 ) ^ { 2 } } { 49 } + \frac { ( y + 5 ) ^ { 2 } } { 40 } = 1$
E) $\frac { ( x + 2 ) ^ { 2 } } { 49 } + \frac { ( y - 5 ) ^ { 2 } } { 40 } = 1$

Question

Find the standard form of the equation of the ellipse with the given characteristics. ​
Foci: $$( - 5 , - 1 ) , ( - 5,5 )$$ ,endpoints of the major axis: $$( - 5 , - 5 ) , ( - 5,9 )$$ ​
A) $\frac { ( x - 5 ) ^ { 2 } } { 40 } + \frac { ( y + 2 ) ^ { 2 } } { 49 } = 1$
B) $\frac { ( x + 5 ) ^ { 2 } } { 40 } + \frac { ( y - 2 ) ^ { 2 } } { 49 } = 1$
C) $\frac { ( x + 5 ) ^ { 2 } } { 49 } + \frac { ( y - 2 ) ^ { 2 } } { 40 } = 1$
D) $\frac { ( x - 2 ) ^ { 2 } } { 49 } + \frac { ( y + 5 ) ^ { 2 } } { 40 } = 1$
E) $\frac { ( x + 2 ) ^ { 2 } } { 49 } + \frac { ( y - 5 ) ^ { 2 } } { 40 } = 1$

Quizplus · Accepted Answer

The center of the ellipse is the midpoint of the foci and the major axis, which is $(-5, 2)$. The length of the major axis is $9 - (-5) = 14$, so $a = 7$ and $a^2 = 49$. The distance between the foci is $5 - (-1) = 6$, so $c = 3$ and $c^2 = 9$. Using $b^2 = a^2 - c^2$, we find $b^2 = 49 - 9 = 40$. The major axis is vertical, so $a^2$ goes with $y$ and $b^2$ with $x$, leading to the equation $\frac{(x + 5)^2}{40} + \frac{(y - 2)^2}{49} = 1$.

Find the Standard Form of the Equation of the Ellipse (−5,−1),(−5,5)( - 5 , - 1 ) , ( - 5,5 )(−5,−1),(−5,5)

Find the Standard Form of the Equation of the Ellipse $( - 5 , - 1 ) , ( - 5,5 )$